Q · 20

Capacitor across a battery — energy supplied

Electrostatics · Capacitance

Question

A capacitor of capacity \(C\) is connected to a battery of potential \(V\). The distance between its plates is reduced to half, the battery remaining connected. The energy given by the battery in this process is:

The Circuit — before & after

Before · gap = d

\(C,\;\; Q_i = CV\)

V d

After · gap = d/2

\(C' = 2C,\;\; Q_f = 2CV\)

V d/2

Battery stays connected ⇒ \(V\) is held fixed. Halving the gap doubles \(C\), which draws extra charge from the battery.

Given

Capacitance\(C\)
Battery EMF\(V\) (stays connected)
New separation\(d \to d/2\)
So \(C \propto 1/d\)\(C' = 2C\)

Asked

WantedEnergy supplied by the battery, \(W_B\)
Key idea\(W_B = V\,\Delta Q\) — work to move charge, not the change in stored energy

Concept

When a battery stays connected, it keeps the plate voltage at \(V\). If the capacitance changes, the charge adjusts to \(Q = CV\), and the battery must push (or pull) that extra charge \(\Delta Q\) across its terminals. The energy it spends doing so is

$$ W_B = V \,\Delta Q = V\,(Q_f - Q_i). $$

Watch the trap: only half of \(W_B\) ends up as extra stored energy; the other half is dissipated as heat in the connecting wires. The question asks for the full energy the battery delivers.

Method & Steps

  1. New capacitance: \(C' = \dfrac{\varepsilon_0 A}{d/2} = 2C.\)
  2. Initial charge: \(Q_i = CV.\)
  3. Final charge (V fixed): \(Q_f = C'V = 2CV.\)
  4. Extra charge drawn: \(\Delta Q = Q_f - Q_i = 2CV - CV = CV.\)
  5. Energy given by battery: \(W_B = V\,\Delta Q = V\cdot CV = \boxed{CV^2}.\)

Easy Trick

Battery connected → think charge, not energy. Capacitance doubles, so charge doubles: \(CV \to 2CV\). The battery only paid for the new charge \(\Delta Q = CV\), at price \(V\) each → \(\;W_B = CV^2\). (If a question instead asked for the rise in stored energy, halve it: \(\tfrac12 CV^2\).)

d
Energy given by the battery \(= CV^2\).
Stored energy rises only \(\tfrac12 CV^2\); the remaining \(\tfrac12 CV^2\) is lost as heat.